Question

Prove that:
$\sin 6°\sin 42°\sin 66°\sin 78°=\frac{1}{16}$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\frac{1}{4}\left(2\sin 6°\sin 66°\right)\left(2\sin 42°\sin 78°\right) \\ =\frac{1}{4}\left(\cos 60°-\cos 72°\right)\left(\cos 36°-\cos 120°\right) \\ \left[2\sin A\sin B=\cos \left(A\mathit{-}B\right)-\cos \left(A+B\right)\right] \\ =\frac{1}{4}\left(\frac{1}{2}-\sin 18°\right)\left(\frac{\sqrt{5}+1}{4}+\frac{1}{2}\right) \end{gathered}$$

$$\begin{gathered} =\frac{1}{4}\left(\frac{1}{2}-\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt{5}+1}{4}+\frac{1}{2}\right) \\ =\frac{1}{4}\left(\frac{2-\sqrt{5}+1}{4}\right)\left(\frac{\sqrt{5}+1+2}{4}\right) \\ =\frac{1}{64}\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right) \\ =\frac{1}{64}\left(9-5\right)=\frac{1}{16} \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$